(3/2)b+5=20-b

Solution for (3/2)b+5=20-b equation:

(3/2)b+5=20-b

We move all terms to the left:

(3/2)b+5-(20-b)=0


Domain of the equation: 2)b!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+3/2)b-(-1b+20)+5=0

We multiply parentheses

3b^2-(-1b+20)+5=0

We get rid of parentheses

3b^2+1b-20+5=0

We add all the numbers together, and all the variables

3b^2+b-15=0

a = 3; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·3·(-15)
Δ = 181
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{181}}{2*3}=\frac{-1-\sqrt{181}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{181}}{2*3}=\frac{-1+\sqrt{181}}{6}
$